Variational iteration method
Reconsider the problem (1), according to the variational iteration
method with Laplace transform; we have the iteration equality [23]
\(L\left[x_{n+1}\right]=L\left[x_{n}\right]-L\left[\int_{0}^{t}\frac{1}{\omega}\sin{\omega\left(t-\xi\right)}\text{dξ}\right]\)
\(=L\left[x_{n}\right]-\frac{1}{\omega}L\left[\sin\text{ωt}\right]\bullet L\left[{\ddot{x}}_{n}\left(t\right)+x_{n}^{3}\left(t\right)+x_{n}^{2}\left(t\right){\ddot{x}}_{n}\left(t\right)\right]\)(19)
where L is the Laplace transform operator.
Assume the initial approximation be
\(x_{0}=Acos\text{ωt}\) (20)
Then we have
\(L\left[x_{1}\left(t\right)\right]=L\left[\text{Acos}\text{ωt}\right]-\frac{1}{\omega}L\left[\sin\text{ωt}\right]\bullet L\left[-A\omega^{2}\cos\omega t+A^{3}\cos^{3}\omega t-A^{3}\omega^{2}\cos^{3}\text{ωt}\right]\)
\(\ =L\left[\text{Acos}\text{ωt}\right]-\frac{1}{\omega}L\left[\sin\text{ωt}\right]L\left[\left(\frac{3\left({{A^{3}-A}^{3}\omega}^{2}\right)}{4}-A\omega^{2}\right)\cos\omega t+\frac{A^{3}-{A^{3}\omega}^{2}}{4}cos3\text{ωt}\right]\)(21)
Imposing the inverse Laplace transform on equality (21), there holds the
first-order approximant
\(x_{1}\left(t\right)=A\cos\text{ωt}-\frac{t}{4\omega}\left(3\left({{A^{3}-A}^{3}\omega}^{2}\right)-4A\omega^{2}\right)\sin\text{ωt}+\frac{A^{3}\omega^{2}-A^{3}}{32\omega^{2}}\left(\cos\text{ωt}-\cos 3\text{ωt}\right)\)(22)
No secular term in the equality (22) requires that
\(3\left({{A^{3}-A}^{3}\omega}^{2}\right)-4A\omega^{2}=0\) (23)
which yields this
\(\omega^{2}=\frac{3A^{2}}{4+3A^{2}}\) (24)
Eq. (24) satisfies two scale extremals.